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CheckiO 是面向初学者和高级程序员的编码游戏，使用 Python 和 JavaScript 解决棘手的挑战和有趣的任务，从而提高你的编码技能，本博客主要记录自己用 Python 在闯关时的做题思路和实现代码，同时也学习学习其他大神写的代码。

CheckiO 官网：https://checkio.org/

我的 CheckiO 主页：https://py.checkio.org/user/TRHX/

CheckiO 题解系列专栏：https://itrhx.blog.csdn.net/category_9536424.html

CheckiO 所有题解源代码：https://github.com/TRHX/Python-CheckiO-Exercise

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题目描述

【Bigger Price】：给定两个参数，第一个为整形，第二个是由字典组成的列表，要求在该列表中，按照字典中 price 关键字的值，查找前几个最大的字典，输出由最大的几个字典组成的列表。

【链接】：https://py.checkio.org/mission/bigger-price/

【输入】：两个参数，整数和字典列表，每个字典都有两个键：name 和 price。

【输出】：由最大的几个字典组成的列表

【范例】：

bigger_price(2, [{"name": "bread", "price": 100},{"name": "wine", "price": 138},{"name": "meat", "price": 15},{"name": "water", "price": 1}
]) == [{"name": "wine", "price": 138},{"name": "bread", "price": 100}
]bigger_price(1, [{"name": "pen", "price": 5},{"name": "whiteboard", "price": 170}
]) == [{"name": "whiteboard", "price": 170}]

解题思路

利用 sorted() 方法对字典按照 price 的值进行排序，此处可以使用匿名函数 lambda，最后按照给定的整数进行切片即可。

匿名函数参考：https://blog.csdn.net/zjuxsl/article/details/79437563

代码实现

def bigger_price(limit: int, data: list) -> list:"""TOP most expensive goods"""return sorted(data, key=lambda i: i["price"], reverse=True)[0:limit]if __name__ == '__main__':from pprint import pprintprint('Example:')pprint(bigger_price(2, [{"name": "bread", "price": 100},{"name": "wine", "price": 138},{"name": "meat", "price": 15},{"name": "water", "price": 1}]))# These "asserts" using for self-checking and not for auto-testingassert bigger_price(2, [{"name": "bread", "price": 100},{"name": "wine", "price": 138},{"name": "meat", "price": 15},{"name": "water", "price": 1}]) == [{"name": "wine", "price": 138},{"name": "bread", "price": 100}], "First"assert bigger_price(1, [{"name": "pen", "price": 5},{"name": "whiteboard", "price": 170}]) == [{"name": "whiteboard", "price": 170}], "Second"print('Done! Looks like it is fine. Go and check it')

大神解答

大神解答 NO.1

import heapqbigger_price = lambda limit, data: heapq.nlargest(limit, data, key=lambda item: item["price"])

大神解答 NO.2

bigger_price = lambda l, d: sorted(d, key=lambda e: -e["price"])[:l]